HYACINTHOS 5429

• Dear Antreas, you wrote:
  
  

  > I uploaded the file 184.doc that Alex sent me.
  > I don't know what it contains, since I don't read MS-Word
  > format files.
  > May someone read it and post a brief description?
  
  

  I difficultly understood (I think) what Alex with his figure
  in the uploaded file declares.
  In a slightly modified figure and in my own words Alex's
  problem is the following:
  In a triangle ABC find the point P with the following
  property: If a, b, c are the sides and k a positive number
  then there are points
  Ab, Ac on BC (Ab closer to B)
  Bc, Ba on CA
  Ca, Cb on AB such that
  PAb = PAc = BaCa = ka
  PBc = PBa = CbAb = kb
  PCa = PCb = AcBc = kc
  i.e. the triangles PAbAc, PBcBa, PCaCb are isosceles
  and the triangles PBaCa, PCbAb, PAcBc are
  congruent between them and antisimilar to ABC.
  
  Very good problem Alex if it is as I understood
  and sorry if I misunderstood.
  
  One solution is: If p, q, r are the angles of the isosceles
  triangles in P then it is easy to prove that
  p = 180 - 4A, q = 180 - 4B, r = 180 - 4C
  and the actual normal x of P is
  x = kacos(90 - 2A) = 2kasinAcosA and hence the
  normals of P are [aacosA : bbcosB : cccosC]
  and this point is X184.
  
  But if the triangle is not acute-angled perhaps it
  doesn't work?
  
  Best regards
  Nikos Dergiades